Impeccable logic

[EdSet100]

Softus,

I agree that the desired box is, 19 times out of 20, not on the player's table and that you could argue that only 1 time in 20 is the desired box on the player's table.

Therefore, the player only has a 1 in 20 chance of winning and that, if the other box has the prize and as we know the other box, being one of the other 19, must represent the other 19 possibilities, that the player must swap.

However, you have not excluded the possibility that this is the 1 in 20 chance in favour of the player. Furthermore, you have to consider that the desired prize has got to where it is (the last box) as a 1 in 19 chance. Therefore both boxes have, against all the odds (20 x 19 to 1), an equal chance of holding the desired prize.

Regarding the Monty Hall puzzle: the web page software is flawed. You will always pick a wrong door due to the software. Therefore you have been pre-steered to a loss. You are then shown another losing door, which had a 1/3 chance of winning. However, because it is always going to be a loser (software) the odds of the 3rd door having the prize must improve from 1 in 3 to 2 in 3 (it has taken the 1/3 from the losing door to balance the odds against the player). This is reflected in the results. Where the web page goes wrong is that you will never pick the winner at the outset. Therefore the winning results are skewed against the non-switch player.

 

[joe-90]

Whatever is in those boxes is in those boxes. It may be 50p or £250K or £100 or £20K.

No - the OP states very clearly that "The boxes have in them £250,000 and £1".

But you fail to take in to account that the game with £250K and £1 will come up every now and again in a random sequence. It makes no difference what the OP states - all permutations of the last 2 boxes will come up randomly.
In this particular case we are talking about, random chance has left us with £250K and £1 - but there is never anything other than random chance involved. Even if the banker knew the content of each box he can only offer a 50-50 swap, whereas Monty Hall had 3 doors to choose from and KNEW which one had the car behind it - so ALWAYS offered one with a goat, which is where the odds change.

 

[Kes]

I have been familiar with the Monty hall paradox for some time, and I believe that the general consensus is, after years of academic wrangling, that swapping is the correct strategy (unless, as someone put it, you want to win the goat). However I pondered over the DonD setup, and wondered whether MH logic applied to this. I still am not sure.

The part that foxes me is that if you have a large number of boxes, say a million and one loaded with values of 1p up to £1m, the odds of picking the highest value are 1,000,000/1. Negligible. Now if Noel, or anyone else, says that he's opened 999,999 boxes and none of them contain the £1m, would you swap?

How does Noel telling you that 999,999 boxes don't have the prize alter the odds? Or does Noel telling you just alter the odds from 1,000,000/1 to 1/2? But you already knew that 999,999 boxes wouldn't hold the prize.

And if that is so, why do the odds in the Monty Hall problem remain at 2/3 to swap after Monty has opened the empty box?

Joe's answer is very persuasive. but would the odds go to 50/50 if Monty opened the door with the goat by chance?

DonD must be one of the most vicious games on TV. It's not just the throw of a dice, or your more astute opponent, or blind chance in any form that snatches the dream away from you. It's a decision that you make that sends you back to your former drudgery and a lifetime of wishful thinking. Just look at their little faces crumbling as they chose the wrong box, or greed takes them one step too far. It's painful. Not that I watch it, of course.

Rgds.

By the way in the OP it's not really £250k or £1, but £250k and any other inferior value. But that doesn't make any difference, does it?

 

[Softus]

However, you have not excluded the possibility that this is the 1 in 20 chance in favour of the player. Furthermore, you have to consider that the desired prize has got to where it is (the last box) as a 1 in 19 chance. Therefore both boxes have, against all the odds (20 x 19 to 1), an equal chance of holding the desired prize.

You have a point, in that I seem to have made an error there. I'll have a think about that.

Regarding the Monty Hall puzzle: the web page software is flawed. You will always pick a wrong door due to the software.

I don't agree - the software allows you to choose the correct door, and to swap to the incorrect door - it happens roughly 1/3 of the time.

 

[tim west]

must admit i won the car first time but i'm not sure the whether the web page accurately depicts the scenario.

 

[joe-90]

If we simplify it is becomes very clear.

20 Boxes. 18 of them empty. 2 contain apples.

Sooner or later via random chance, a contestant will end up with two boxes that both have apples in them. One on the table and one out on the counter. It matters not whether Noel or the Banker know what is in the boxes. They both contain apples - so switching the boxes is a complete irrelevance.

Now dya gedditt?

 

[tim west]

the difference is one apple is worth more than the other, but I get your gist.
I still believe that the producers can influence the choice though.

 

[Kes]

If we simplify it is becomes very clear.

20 Boxes. 18 of them empty. 2 contain apples.

Sooner or later via random chance, a contestant will end up with two boxes that both have apples in them. One on the table and one out on the counter. It matters not whether Noel or the Banker know what is in the boxes. They both contain apples - so switching the boxes is a complete irrelevance.

Now dya gedditt?

Well, I simplyfied it and came to a different conclusion.

You're looking for the Ace of Spades. Deal yourself one card, and 51 to your imaginary opponent (imaginary, because by now no-one in the family will have anything to do with you). Your opponent now throws away 50 of his cards that aren't the Ace of Spades. So it's one to one.

So far I've lost every time I've played, unless I swap, whan I've won every time. If I play enough times I will win without swapping, but not often.

The hypothetical DonD contestant is not facing one box, but 20, of which 19 containing the wrong value have been discarded.

 

[joe-90]

There are either apples in the box or there aren't. Argue until you are blue in the face but the fact remains it is nothing but random chance.
Forget your 'wrong values', you are arguing from both sides of the equation and getting nonsense.

 

[joe-90]

the difference is one apple is worth more than the other, but I get your gist.
I still believe that the producers can influence the choice though.

But if they are both Granny Smiths - so what?

 

[Trazor]

Well, I simplyfied it and came to a different conclusion.

You're looking for the Ace of Spades. Deal yourself one card, and 51 to your imaginary opponent (imaginary, because by now no-one in the family will have anything to do with you). Your opponent now throws away 50 of his cards that aren't the Ace of Spades. So it's one to one.

So far I've lost every time I've played, unless I swap, whan I've won every time. If I play enough times I will win without swapping, but not often.

The hypothetical DonD contestant is not facing one box, but 20, of which 19 containing the wrong value have been discarded.

I think you have misunderstood the scenario here......

Its a totally false analogy, as your opponent will hold the Ace of spades on average, 51 times out of every 52 games, therefore you must swap.

 

[tim west]

I cant work out how having more than the original three choices could be seen as simplifying things?

 

[JohnD]

I understood it.

 

[joe-90]

I cant work out how having more than the original three choices could be seen as simplifying things?

There are only 2 choices. There is no third choice that the game-show host could know and thus could not influence the outcome.

 

[JohnD]

Joe didn't understand it.